An x-rays tube takes a current of 5mA and operates at a potential difference of 50kV. what power is dissipated?
Answers
Answered by
0
Explanation:
(a) Heat produced per second at the target
P=0.995 VI(0.5% of energy is converted into x−rays)
⇒ I=
0.995 V
P
=
0.995 V
720
=
0.995×40×10
3
720
=0.018 ampere
The no. of e−emitted per second :
n=
e
I
=
1.6×10
−19
0.018
=1.1×10
17
electron
(b) Energy of photon
2
1
mv
2
=eV
⇒ v=
m
2 eV
=1.2×10
8
m/s
Answered by
0
Answer:
(a) Heat produced per second at the target
P=0.995 VI(0.5% of energy is converted into x−rays)
⇒ I=
0.995 V
P
=
0.995 V
720
=
0.995×40×10
3
720
=0.018 ampere
The no. of e−emitted per second :
n=
e
I
=
1.6×10
−19
0.018
=1.1×10
17
electron
(b) Energy of photon
2
1
mv
2
=eV
⇒ v=
m
2 eV
=1.2×10
8
m/s
Similar questions